Question

A man is moving away from a tower 41.6 m high at the rate of 2 m/sec. Find the rate at which the angle of elevation of the top of the tower is changing, when he is at a distance of 30 m from the foot of the tower. Assume that the eye level of the man is 1.6m from the ground.

Answer 1

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$\tan\alpha = 40 \div 30 \\ \alpha = \tan {}^{ - 1} ( 4\div 3)$
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